Match List-I with List-II.

List-I (Conversion)	List-II (Reagents, Conditions used) [cite: 248, 249]
(A) Chlorobenzene arrow$\rightarrow$ Phenol	(I) Warm, H₂O$\text{Warm, } \text{H}_2\text{O}$
(B) p-Nitrochlorobenzene arrow$\rightarrow$ p-Nitrophenol	(II) (a) NaOH, 368 K$\text{NaOH, } 368\text{ K}$; (b) H3O^+$\text{H}3\text{O}^+$	(C) 2,4-Dinitrochlorobenzene arrow$\rightarrow$ 2,4-Dinitrophenol	(III) (a) NaOH, 443 K$\text{NaOH, } 443\text{ K}$; (b) H₃O^+$\text{H}_3\text{O}^+$	(D) 2,4,6-Trinitrochlorobenzene arrow$\rightarrow$ 2,4,6-Trinitrophenol	(IV) (a) NaOH, 623 K, 300 atm$\text{NaOH, } 623\text{ K, } 300\text{ atm}$; (b) H₃O^+$\text{H}_3\text{O}^+$ Choose the correct answer from the options given below: A. (A)-(II), (B)-(III), (C)-(I), (D)-(IV)$\text{(A)-(II), (B)-(III), (C)-(I), (D)-(IV)}$ B. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)$\text{(A)-(III), (B)-(IV), (C)-(II), (D)-(I)}$ C. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)$\text{(A)-(IV), (B)-(III), (C)-(II), (D)-(I)}$ D. (A)-(IV), (B)-(III), (C)-(I), (D)-(II)$\text{(A)-(IV), (B)-(III), (C)-(I), (D)-(II)}$ Solution & Explanation Related Formula Rate of SNAr ∝ Number of electron-withdrawing groups (-I, -M) at ortho/para positions$$\text{Rate of } S_N\text{Ar} \propto \text{Number of electron-withdrawing groups (-I, -M) at ortho/para positions} $$ Core Logic Aryl halides are generally unreactive towards nucleophilic substitution due to resonance stabilization of the C-Cl$\text{C-Cl}$ bond. However, the presence of strong electron-withdrawing groups (-NO₂$-\text{NO}_2$) at ortho and para positions dramatically increases reactivity by stabilizing the intermediate carbanion: - (A) Chlorobenzene: Needs extreme conditions: NaOH at 623 K, 300 atm$\text{NaOH at } 623\text{ K, } 300\text{ atm}$ (Dow's Process) arrow$\rightarrow$ (IV) - (B) p-Nitrochlorobenzene: One para -NO₂$-\text{NO}_2$ group softens required temperature to 443 K$443\text{ K}$ arrow$\rightarrow$ (III) - (C) 2,4-Dinitrochlorobenzene: Two electron-withdrawing groups lower needed temperature further to 368 K$368\text{ K}$ arrow$\rightarrow$ (II) - (D) 2,4,6-Trinitrochlorobenzene: Highly activated picryl chloride hydrolyzes smoothly with just warm water arrow$\rightarrow$ (I) Step 1: Final Match Alignment Matching sequences cleanly yields: (A)-(IV), (B)-(III), (C)-(II), (D)-(I). Pattern Recognition The more -NO₂$-\text{NO}_2$ groups present on the ring, the less aggressive the reagent/temperature setup required. Count -NO₂$-\text{NO}_2$ groups: 0 arrow 623K$0 \rightarrow 623\text{K}$, 1 arrow 443K$1 \rightarrow 443\text{K}$, 2 arrow 368K$2 \rightarrow 368\text{K}$, 3 arrow warm water$3 \rightarrow \text{warm water}$. Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Reference Study Guides Haloalkanes And Haloarenes Keywords: #Dows process chlorobenzene #JEE Main 2025 Evening Q32 #nucleophilic aromatic substitution kinetics #picryl chloride hydrolysis More Haloalkanes and Haloarenes Previous-Year Questions Q54 jee_main_2026_21_jan_morning Reactions of Haloalkanes A hydrocarbon ‘P’ (C₄H₈$C_{4}H_{8}$) on reaction with HCl gives an optically active compound ‘Q’ (C₄H₉Cl$C_{4}H_{9}Cl$) which on reaction with one mole of ammonia gives compound ‘R’ (C₄H₁₁N$C_{4}H_{11}N$). ‘R’ on diazotization followed by hydrolysis gives ‘S’. Identify P, Q, R and S. A. P = CH₃ - CH₂ - CH = CH₂, Q = CH₃ - CH₂ - CH₂ - CH₂Cl, R = CH₃ - CH₂ - CH₂ - NH₂, S = CH₃ - CH₂ - CH(OH) - CH₃$\mathrm{P = CH_3 - CH_2 - CH = CH_2}, \mathrm{Q = CH_3 - CH_2 - CH_2 - CH_2Cl}, \mathrm{R = CH_3 - CH_2 - CH_2 - NH_2}, \mathrm{S = CH_3 - CH_2 - CH(OH) - CH_3}$ B. P = Cyclobutane, Q = 1-Chlorobutane, R = Butan-1-amine, S = Cyclobutanol$\mathrm{P = Cyclobutane}, \mathrm{Q = 1-Chlorobutane}, \mathrm{R = Butan-1-amine}, \mathrm{S = Cyclobutanol}$ C. P = CH₃ - CH = CH - CH₃, Q = CH₃ - CH₂ - CH(Cl) - CH₃, R = CH₃ - CH₂ - CH(NH₂) - CH₃, S = CH₃ - CH₂ - CH(OH) - CH₃$\mathrm{P = CH_3 - CH = CH - CH_3}, \mathrm{Q = CH_3 - CH_2 - CH(Cl) - CH_3}, \mathrm{R = CH_3 - CH_2 - CH(NH_2) - CH_3}, \mathrm{S = CH_3 - CH_2 - CH(OH) - CH_3}$ D. P = CH₃ - CH = CH - CH₃, Q = CH₃ - CH₂ - CH₂ - CH₂ - Cl, R = CH₃ - CH₂ - CH₂ - CH₂ - NH₂, S = CH₃ - CH₂ - CH₂ - CH₂ - OH$\mathrm{P = CH_3 - CH = CH - CH_3}, \mathrm{Q = CH_3 - CH_2 - CH_2 - CH_2 - Cl}, \mathrm{R = CH_3 - CH_2 - CH_2 - CH_2 - NH_2}, \mathrm{S = CH_3 - CH_2 - CH_2 - CH_2 - OH}$ Solution Core Logic Since P (C₄H₈$C_4H_8$) reacts with HCl to give an optically active compound Q (C₄H₉Cl$C_4H_9Cl$), P must be But-2-ene. Addition of HCl yields 2-chlorobutane, which possesses a chiral center. CH₃-CH=CH-CH₃ HCl CH₃-CH₂-C^*H(Cl)-CH₃ (P is But-2-ene, Q is 2-chlorobutane)$$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3 \xrightarrow{\mathrm{HCl}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C}^*\mathrm{H}(\mathrm{Cl})-\mathrm{CH}_3 \quad \text{(P is But-2-ene, Q is 2-chlorobutane)}$$ Compound Q reacts with NH₃$\mathrm{NH}_3$ to undergo nucleophilic substitution forming a primary amine R: CH₃-CH₂-CH(Cl)-CH₃ NH₃ CH₃-CH₂-CH(NH₂)-CH₃ (R is Butan-2-amine)$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{Cl})-\mathrm{CH}_3 \xrightarrow{\mathrm{NH}_3} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3 \quad \text{(R is Butan-2-amine)}$$ Diazotization of primary aliphatic amines followed by hydrolysis yields an alcohol via a carbocation intermediate (S): CH₃-CH₂-CH(NH₂)-CH₃ NaNO₂ / HCl / H₂O CH₃-CH₂-CH(OH)-CH₃ (S is Butan-2-ol)$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3 \xrightarrow{\mathrm{NaNO}_2 / \mathrm{HCl} / \mathrm{H}_2\mathrm{O}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_3 \quad \text{(S is Butan-2-ol)}$$ Pattern Recognition An optically active alkyl halide formed from a C₄H₈$C_4H_8$ alkene with HX is a classic pointer to 2-halobutane originating from But-2-ene or But-1-ene. Diazotization of primary aliphatic amines (R-NH₂$R-NH_2$) with NaNO₂/HCl$NaNO_2/HCl$ yields alcohols (R-OH$R-OH$) with possible rearrangements, though here a secondary carbocation is already stable. Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Amines Q59 jee_main_2026_21_jan_evening Nucleophilic Substitution ($S_N1$) and Carbocation Stability The correct order of reactivity of the following benzyl halides towards reaction with KCN$\text{KCN}$ is: A. (1) a > b > c > d$(1) \ a > b > c > d$ B. (2) b > a > d > c$(2) \ b > a > d > c$ C. (3) b > a > c > d$(3) \ b > a > c > d$ D. (4) a > b > d > d$(4) \ a > b > d > d$ Solution Core Logic The reaction proceeds via an SN1$S_N1$ mechanism for activated benzyl halides or nucleophilic substitution rate depends on carbocation stability / electronic effects of substituents (-OH, -NH₂$-\text{NH}_2$, -NO₂$-\text{NO}_2$, etc.). Amino and hydroxy substituents strongly activate through +M effect. Nitro groups strongly deactivate through -M effect. Step 1: Final Conclusion The correct order of reactivity is b > a > d > c$b > a > d > c$, corresponding to option (2). Pattern Recognition Sees: benzyl halide reactivity with cyanide. Trap: Confusing polar/inductive effects with resonance effects of substituents. Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Q65 jee_main_2026_21_jan_evening Optical Isomerism and Composition Percentage Given below are four compounds: (a) n-propyl chloride (b) iso-propyl chloride (c) sec-butyl chloride (d) neo-pentyl chloride Percentage of carbon in the one which exhibits optical isomerism is: A. (1) 52$(1) \ 52$ B. (2) 56$(2) \ 56$ C. (3) 46$(3) \ 46$ D. (4) 40$(4) \ 40$ Solution Core Logic Among the given compounds, sec-butyl chloride (2-chlorobutane) is optically active and contains a chiral center. Molecular formula of 2-chlorobutane arrow C₄H₉Cl$\rightarrow \text{C}_4\text{H}_9\text{Cl}$. Molar mass = 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 g/mol$= 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 \text{ g/mol}$. Step 1: Calculating Percentage of Carbon % of C = (48)/(92.5) × 100 = 51.89% ≈ 52%$$\text{\% of C} = \frac{48}{92.5} \times 100 = 51.89\% \approx 52\%$$ Pattern Recognition Sees: optical isomerism identification combined with elemental percentage composition calculation. Trap: Selecting the wrong halogen derivative for optical activity. Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Q59 jee_main_2026_22_january_morning Nucleophilicity Trends The correct order of reactivity of CH₃Br$CH_{3}Br$ in methanol with the following nucleophiles is F⁻$F^{-}$, I⁻$I^{-}$, C₂H₅O⁻$C_{2}H_{5}O^{-}$ and C₆H₅O⁻$C_{6}H_{5}O^{-}$ A. I⁻ > C₆H₅O⁻ > F⁻ > C₂H₅O⁻$I^{-} > C_{6}H_{5}O^{-} > F^{-} > C_{2}H_{5}O^{-}$ B. I⁻ > C₂H₅O⁻ > C₆H₅O⁻ > F⁻$I^{-} > C_{2}H_{5}O^{-} > C_{6}H_{5}O^{-} > F^{-}$ C. I⁻ > C₂H₅O⁻ > F⁻ > C₆H₅O⁻$I^{-} > C_{2}H_{5}O^{-} > F^{-} > C_{6}H_{5}O^{-}$ D. I⁻ > F⁻ > C₆H₅O⁻ > C₂H₅O⁻$I^{-} > F^{-} > C_{6}H_{5}O^{-} > C_{2}H_{5}O^{-}$ Solution Related Formula Rate ∝ Nucleophilicity$$Rate \propto \text{Nucleophilicity}$$ Core Logic The reaction of CH₃Br$CH_3Br$ (a primary halide) occurs via the SN2$S_N2$ mechanism. The rate depends directly on the nucleophilicity of the attacking species. Methanol is a polar protic solvent. In polar protic solvents, larger halide ions are better nucleophiles because they are less solvated. Therefore, I^-$I^-$ is a stronger nucleophile than F^-$F^-$. For the alkoxide and phenoxide, C₂H₅O^-$C_2H_5O^-$ is a stronger nucleophile than C₆H₅O^-$C_6H_5O^-$ because the negative charge on phenoxide is delocalized over the benzene ring, reducing its electron-donating ability. Step 1: Final Order Combining these factors, the overall order of nucleophilicity in methanol is: I⁻ > C₂H₅O⁻ > C₆H₅O⁻ > F⁻$$I^{-} > C_{2}H_{5}O^{-} > C_{6}H_{5}O^{-} > F^{-}$$ Pattern Recognition In polar protic solvents: Size dominates (down a group, nucleophilicity increases). For species of similar size, less resonance stabilization means stronger nucleophile. Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Q67 jee_main_2026_22_january_morning SN1 Mechanism The correct order of the rate of reaction of the following reactants with nucleophile by SN1$S_{N}1$ mechanism is: (Given: Structure I and II are rigid) A. IV < III < II < I$IV < III < II < I$ B. III < I < II < IV$III < I < II < IV$ C. II < I < III < IV$II < I < III < IV$ D. I < II < III < IV$I < II < III < IV$ Solution Related Formula Rate of SN1 ∝ Stability of intermediate carbocation$$\text{Rate of } S_N1 \propto \text{Stability of intermediate carbocation}$$ Core Logic Evaluate the stability of the carbocations formed upon dissociation of the bromide ion: Structure (I) and (II) form carbocations at bridgehead positions of rigid bicyclic systems. According to Bredt's rule, these are highly unstable because they cannot adopt the necessary planar sp²$sp^2$ geometry. Among them, (I) has an additional methyl group which provides slight +I stabilization compared to (II). So, (II) is even less stable than (I). Structure (III) forms a tertiary carbocation, but it's not a rigid bridgehead preventing planarity, so it's significantly more stable than I and II. Structure (IV) is a highly stable trityl-like carbocation where the positive charge is stabilized by extensive resonance with the adjacent phenyl ring(s). Step 1: Final Ordering Stability order: (II) < (I) < (III) < (IV)$(II) < (I) < (III) < (IV)$. Thus, the SN1$S_{N}1$ rate follows the same order. Pattern Recognition Bridgehead halides effectively do not undergo SN1$S_N1$ (or SN2$S_N2$) reactions due to Bredt's rule. Resonance stabilized carbocations always dominate. Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes More Haloalkanes and Haloarenes Questions — jee_main_2025_07_april_evening Practice all Haloalkanes and Haloarenes previous-year questions → The major product of the following reaction is: Given below are two statements : Statement I: will undergo Which among the following halides will generate the most stable The product B formed in the following reaction sequence is The products A and B in the following reactions, respectively Given below are two statements: Statement (I): is more polar The structure of the major product formed in the following Given below are two statements : Statement-I: The conversion proceeds Consider the following sequence of reactions: Consider the above sequence A hydrocarbon ‘P’ ( ) on reaction with HCl gives Which of the following reaction is NOT correctly represented ? Two products X and Y are formed in the following The correct order of reactivity of the following benzyl halides The correct order of reactivity of in methanol with the Match the List-I with List-II <div style="overflow-x: auto; margin: 1rem Given below are two statements: Statement I: High concentration of Given below are four compounds: (a) n-propyl chloride (b) iso-propyl The correct order of the rate of reaction of the Given below are two statements: Statement I : bond is As compared with chlorocyclohexane, which of the following statements correctly The dibromo compound [P] (molecular formula: ) when heated with Given below are two statement one is labeled as Assertion Consider all the structural isomers with molecular formula are separately The correct statement regarding nucleophilic substitution reaction in a chiral The total number of hydrogen atoms in product A and SYSTEM Syllabus Analysis & Trend Mapping Dashboard Chapter Trends Formula Recognition Questions Feed Weightage Login Sign Up Dashboard Rankbit System Revise Mistakes Rankbit JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) Report Issue Question Issue Type Wrong Answer Wrong Solution Formula Error Typo / Language Issue Diagram Issue Option Mismatch Duplicate Question Out of Syllabus Other Optional Notes Cancel Submit Report Quick Review Question Accuracy Feedback Accurate Needs Fix Difficulty Feedback Easy Medium Hard Confusion Score Low High Optional Notes Cancel Submit Review Welcome back Continue with Google or Sign In Don't have an account? Sign up Enter the 6-digit code we emailed to your email Verify & Continue Didn't get it? Resend code ← Back Enter your account email and we'll send you a link to reset your password. Send reset link ← Back to sign in